measurable sets not depending on even coordinates - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T08:22:44Z http://mathoverflow.net/feeds/question/8166 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8166/measurable-sets-not-depending-on-even-coordinates measurable sets not depending on even coordinates Ori Gurel-Gurevich 2009-12-08T05:47:56Z 2010-06-09T07:22:16Z <p>Let $A\subset\{0,1\}^\omega$ be a measurable set (w.r.t. the usual borel sigma algebra) which does not depend on any even coordinate (that is, if $x\in A$ and $x$ and $y$ agree except on a finite number of even coordinates, then $y\in A$).</p> <p>Is it true that $A$ belongs to the sigma-algebra generated by all the odd coordinates + the tail sigma algebra?</p> <p>To clarify: the tail sigma algebra consists of all the events which do not depend on any coordinate.</p> <p>It seems to me that this should be some easy/well known measure theory fact/counterexample, but perhaps I'm wrong? Suggestions on where to look for an answer would be welcome.</p> <p>Note that it is well known the this statement is false if the ground space would be $[0,1]^\omega$.</p> http://mathoverflow.net/questions/8166/measurable-sets-not-depending-on-even-coordinates/8186#8186 Answer by Gerald Edgar for measurable sets not depending on even coordinates Gerald Edgar 2009-12-08T13:42:45Z 2009-12-08T13:50:34Z <p>Write $\{0,1\}^\mathbb{N}$ as cartesian product $\{0,1\}^E \times \{0,1\}^O$, where $E$ is the evens and $O$ is the odds. Your usual sigma-algebra is the product sigma-algebra that goes with this product.</p> <p>So we need something like this: Write $\otimes$ for product sigma-algebra. Then is $$\mathcal{F}\otimes\left(\bigcap_{k=1}^\infty\mathcal{G}_k\right) = \bigcap_{k=1}^\infty\left(\mathcal{F}\otimes\mathcal{G}_k\right)$$</p>